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scoray [572]
3 years ago
10

Blow up a balloon and rub it against your shirt a number of times. In doing so you give the balloon a net electric charge. Now t

ouch the balloon to the ceiling. On being released, the balloon will remain stuck to the ceiling. Why?
Physics
1 answer:
tamaranim1 [39]3 years ago
8 0
<span>Balloons are blown up, and then rubbed against your shirt many times. The balloon then touches the ceiling. When released, the balloon remains stuck to the ceiling. The balloon is charged by contact. The ceiling has a neutral charge. The charged balloon induces a slight surface charge on the ceiling opposite to the charge on the balloon. Balloon and ceiling electric charges are opposite in sign, so they will attract each other. Since both the balloon and the ceiling are insulators, charge can not flow from one to the other. The charge on the balloon is fixed on the balloon and the charge on the ceiling remains fixed to the ceiling. It just so happens that the<span> electrostatic force the ceiling exerts on the balloon is sufficient to hold the balloon in place (i.e. overcomes gravity, etc.).</span></span>
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Anaya Goodloe decides to spike the football after a big touchdown (on Earth) and it
Llana [10]

Answer:

The football leaves with the velocity, u = 15.68 m/s

Explanation:

Given data,

The football bounces back up off the ground and is airborne for, t = 3.2 s

Let the football bounces back up off the ground in the vertical direction

The formula for time of flight is given by,

                              t = 2u /g

∴                             u = gt / 2

Substituting the values,

                               u = 9.8 x 3.2 / 2

                                u = 15.68 m/s

Hence, the football leaves with the velocity, u = 15.68 m/s

6 0
3 years ago
What does a physicist study?
MissTica

Answer:

A. Matter and energy

Explanation:

5 0
3 years ago
Calculate the elastic potential energy stored in a spring if it has a force constant of 150 N/m. the spring is extended to a len
Alenkinab [10]

Answer:

6.75J

Explanation:

U=1/2KΔx²

U=0.5* 150*0.30^2

4 0
2 years ago
The angle a vector makes with the x-axis of a coordinate system must be what size in order to make one or more of its components
Aneli [31]
If only 1 option is correct then it is (D)
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3 0
3 years ago
Read 2 more answers
A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A pie
liubo4ka [24]

Answer:

Explanation:

Capacitance of the capacitor = 13.5μF

Voltage across plate is 24V

Dielectric constant k=3.55.

a. Energy in capacitor is given by

E=1/2CV^2

We want to calculate energy without the dielectric substance

Given that C=13.5 μF and V=24V

The capacitance give is with dielectric so we need to remove it

C=kCo

Co=C/k

Then the Co=13.5μF/3.55

Co=3.803μF

Then

E=(1/2)×3.803×10^-6×24^2

E=1.1×10^-3J

E=1.1mJ

b. Energy in capacitor is given by

E=1/2CV^2

The capacitance given is with a dielectric, so we are going to apply it direct.

Given that C=13.5 μF and V=24V

Then

E=(1/2)×13.5×10^-6×24^2

E=3.89×10^-3J

E=3.9mJ

c. The energy without dielectric is 1.1mJ and the energy with dielectric is 3.9mJ

The energy increase when the dielectric material is added

d. Dielectrics in capacitors serve three purposes: to keep the conducting plates from coming in contact, allowing for smaller plate separations and therefore higher capacitances;

Therefore, Since dielectric allow higher capacitance, and energy of a capacitor is directly proportional to the capacitance, then the higher the capacitance the higher the energy.

6 0
3 years ago
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